# The speed of a projectile when it reaches its maximum height

1. May 21, 2016

### Ab17

1. The problem statement, all variables and given/known data
The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi- mum height. What is the initial projection angle of the projectile?

2. Relevant equations
Vf2 = vi2 + 2ax

3. The attempt at a solution
I know that the speed at the top only consist of an x component and that the speed at half the height would be sqrt(vx2+vy2) . But im trying everything from two hours and not getting a solution. The answer is suppost to be 67.8 but I dont know how to get it someone help

2. May 21, 2016

### Mastermind01

The max height of the projectile is given by $\frac {u^2sin^2(\theta)}{2g}$ where $\theta$ is the angle of projection. Half the height would be $\frac {u^2sin^2(\theta)}{4g}$ . Find the y-component of velocity at half the height . Find the speed by adding the x and y components. Equate it to twice the x-component. Solve for $\theta$ . You should get your answer.

3. May 21, 2016

### Ab17

X comp : Vicos@t
Y comp : sqrt(visin@ -2gh/2)

4. May 21, 2016

### Ab17

X comp : Vicos@ ***

5. May 21, 2016

### Ab17

Vf = sqrt((vicos@)^2 + Visin@-gh)

6. May 22, 2016

### Ab17

Are those the correct components?

7. May 22, 2016

### Mastermind01

The x-component is correct ($v_icos (\theta)$) the y-component is wrong. You should learn $\LaTeX$ else your equations are hard to read.

8. May 22, 2016

### Ab17

Visin@ - gt ?

9. May 23, 2016

### Mastermind01

You are just giving me equations, I don't think you are putting in much effort. For the y-component $v_i = v_isin(\theta)$ , acceleration $a = -g$ and height $h = \frac{v_i^2sin^2(\theta)}{4g}$ What equation of motion do you use now to get $v_f$ ?

10. May 23, 2016

### Ab17

Vf2 = vi2 + 2ax

11. May 23, 2016

### Ab17

So I will get vi2sin2@/2 for vfy and then i get vf which will be sqrt(vi2cos2@ + vi2sin2@) i equate this to 2

12. May 23, 2016

### Ab17

2viCos@ which gives me Cos@2 = 1/3 which gives an angle of 54.74 while the answer is 67.8

13. May 23, 2016

### Mastermind01

You have clearly made a calculation mistake , using the equation you'll get $v_f^2 = \frac{v_i^2sin^2(\theta)}{2}$ . The resulting equation will be $$4v_i^2cos^2(\theta) = v_i^2cos^2(\theta) + \frac{v_i^2sin^2(\theta)}{2}$$ Solve for $\theta$ you'll get your answer.